Question

The starting salaries of new graduates in USA may differ according to the region that they are living. A group of firms is randomly selected from each region to test this clam. Use the following data to test this claim at 1% significance level. Show your work and do not forget to interpret the result. Note that the data can be transformed to make the computations more reasonable (example: $42,500 = 4.25). North 40500 41500 40000 41000 41500 South 51000 49500 49000 48000 East 45500 43500 45000 46500 46000 West 37000 38500 40000 41500

Answer 1

Group 1 Group 2 Group 3 Group 4 40500 51000 45500 37000 41500 49500 43500 38500 40000 49000 45000 40000 41000 48000 46500 41500 41500 46000 Sum = 204500 197500 226500 157000 Average = 40900 49375 45300 39250 [; x = 8365750000 9756250000 10265750000 6173500000 St. Dev. = 651.92 1250 1151.086 1936.492 SS = 1700000 4687500 5300000 11250000 n = 5 4 5 4 The total sample size is N = 18. Therefore, the total degrees of freedom are: dftotal 18 – 1= 17

Also, the between-groups degrees of freedom are dfbetween = 4-1 =3, and the within-groups degrees of freedom are: dfwithin = d ftotal – dfbetween = 17 – 3 = 14 First, we need to compute the total sum of values and the grand mean. The following is obtained Xij = 204500 + 197500 + 226500 + 157000 = 785500 Also, the sum of squared values is x}} = 8365750000 + 9756250000 + 10265750000+ 6173500000 = 34561250000 Based on the above calculations, the total sum of squares is computed as follows 2 S Stotal = {x};- † ( Xij = 34561250000 7855002 18 = 282902777.778 The within sum of squares is computed as shown in the calculation below: S Swithin = sswithingroups = 1700000 + 4687500 + 5300000 + 11250000 = 22937500 The between sum of squares is computed directly as shown in the calculation below: The between sum of squares is computed directly as shown in the calculation below: Now that sum of squares are computed, we can proceed with computing the mean sum of squares: M Sbetween SSbetween dfbetween 259965277.778 3 = 86655092.593 M Swithin = S Swithin df within 22937500 14 = 1638392.857

Finally, with having already calculated the mean sum of squares, the F-statistic is computed as follows: F= M Sbetween M Swithin 86655092.593 1638392.857 = 52.89

(1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: M1 = M2 = M3 = M4 Ha: Not all means are equal The above hypotheses will be tested using an F-ratio for a One-Way ANOVA. (2) Rejection Region Based on the information provided, the significance level is a = 0.01, and the degrees of freedom are dfi = 3 and df2 = 3, therefore, the rejection region for this F-test is R={F:F> Fc = 5.564} (3) Test Statistics F= M Sbetween M Swithin 86655092.593 1638392.857 52.89 (4) Decision about the null hypothesis Since it is observed that F = 52.89 > Fc = 5.564, it is then concluded that the null hypothesis is rejected. Using the P-value approach: The p-value is p= 0, and since p=0 <0.01, it is concluded that the null hypothesis is rejected. (5) Conclusion It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that not all 4 population means are equal, at the a = 0.01 significance level.

CONCLUSION : there is significant difference at 1% level of significance

(please UPVOTE)